Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
void _copy(int **matrix, int **mas, int n, int m) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) mas[i][j] = matrix[i][j];
}
bool set_col(int **mas, int n, int j, int v) {
  for (int i = 0; i < n; i++)
    if ((mas[i][j] -= v) < 0) return false;
  return true;
}
bool set_row(int **mas, int i, int m, int v) {
  for (int j = 0; j < m; j++)
    if ((mas[i][j] -= v) < 0) return false;
  return true;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int **matrix = new int *[n], **mas = new int *[n], sum = 0, _sum;
  for (int i = 0; i < n; i++) {
    matrix[i] = new int[m];
    mas[i] = new int[m];
    for (int j = 0; j < m; j++) {
      scanf("%d", &matrix[i][j]);
      sum += matrix[i][j];
    }
  }
  std::vector<int> type;
  std::vector<int> numb;
  int ans = 1000000000, now;
  std::vector<int> _type;
  std::vector<int> _numb;
  for (int k = 0; k <= matrix[0][0]; k++) {
    _sum = sum;
    _copy(matrix, mas, n, m);
    now = matrix[0][0];
    _type.clear();
    _numb.clear();
    for (int i = 0; i < k; i++) {
      _type.push_back(1);
      _numb.push_back(1);
    }
    for (int i = 0; i < (matrix[0][0] - k); i++) {
      _type.push_back(2);
      _numb.push_back(1);
    }
    _sum -= k * m;
    if (!set_row(mas, 0, m, k)) continue;
    _sum -= (matrix[0][0] - k) * n;
    if (!set_col(mas, n, 0, matrix[0][0] - k)) continue;
    bool flag = false;
    for (int i = 1; i < n; i++)
      if (mas[i][0]) {
        _sum -= mas[i][0] * m;
        now += mas[i][0];
        for (int c = 0; c < mas[i][0]; c++) {
          _type.push_back(1);
          _numb.push_back(i + 1);
        }
        if (!set_row(mas, i, m, mas[i][0])) {
          flag = true;
          break;
        }
      }
    if (flag) continue;
    for (int j = 1; j < m; j++)
      if (mas[0][j]) {
        _sum -= mas[0][j] * n;
        now += mas[0][j];
        for (int c = 0; c < mas[0][j]; c++) {
          _type.push_back(2);
          _numb.push_back(j + 1);
        }
        if (!set_col(mas, n, j, mas[0][j])) {
          flag = true;
          break;
        }
      }
    if (flag) continue;
    if (!_sum && now < ans) {
      ans = now;
      type = _type;
      numb = _numb;
    }
  }
  if (ans == 1000000000)
    printf("-1");
  else {
    printf("%d", ans);
    for (int i = 0; i < ans; i++)
      if (type[i] == 1)
        printf("\nrow %d", numb[i]);
      else
        printf("\ncol %d", numb[i]);
  }
  return 0;
}